We already have a few questions and answers about the qualitative effects of altitude on fuel economy:

I am interested in understanding how the changes in pressure and temperature at different altitudes affect the fuel economy of a turbofan powered aircraft quantitatively. In the end, I would like to make a plot of relative fuel economy vs. altitude that takes all of these effects into account, but I don't know how to quantitatively combine these effects.

Some notes:

  • By fuel economy I mean fuel required per distance traveled, not time.
  • I am not interested in absolute numbers for the fuel per distance, which would require specifying a particular aircraft. I am only interested in how the fuel per distance figure would relatively change with altitude, e.g. normalized to 1 at sea level.
  • I assume ISA (International Standard Atmosphere) profiles for pressure and altitude.
  • I assume the no wind case. Different winds at different altitudes will of course have an effect on the result, but it is easy to take this into account after the no wind case is understood.
  • Let's assume a typical climb profile for a short- to medium-haul jet airliner: 250/280/0.78

    TAS and Mach for a typical climb profile

    You can see that the TAS increases until reaching Mach 0.78, then decreases due to the lower temperatures causing a lower speed of sound and then remains constant above the tropopause. I am particularly interested in how the fuel economy will behave around these altitudes.

  • 1
    $\begingroup$ Fuel economy per hour, per mile? Set speed or set AoA? A lot depends on gross weight. My flight planning cruise tables for a 737-400 indicate that best economy is found around 2000-5000 feet below service ceiling when operating m.74-m.78. $\endgroup$
    – Max Power
    Commented Sep 23, 2020 at 2:12
  • $\begingroup$ @MaxPower It's in the question: "By fuel economy I mean fuel required per distance traveled, not time." and "Let's assume a typical climb profile for a short- to medium-haul jet airliner: 250/280/0.78" (that is 250KIAS below FL100, 280KIAS above, until reaching Mach 0.78). $\endgroup$
    – Bianfable
    Commented Sep 23, 2020 at 7:05
  • 1
    $\begingroup$ Sorry, I missed that. You still have many issues controlled by the specifics of each aircraft and even loading of each aircraft. You specify mach.78 but some jets have a critical mach below .78, and ground speed will change with temperature even at a set mach. Engines can be designed for any altitude. Similar for wing loading. $\endgroup$
    – Max Power
    Commented Oct 14, 2020 at 1:58

2 Answers 2


I am adding this community wiki answer to show my current state of research and to provide the plot vs. altitude, which I will update when I learn more. Comments are welcome.

Thermal Efficiency

From this answer by Peter Kämpf, we know that the thermal efficiency for jet engines is given by

$$ \eta = \frac{T_\text{max} - T_\text{amb}}{T_\text{max}} $$

where $ T_\text{amb} $ is just the ambient temperature (from ISA) and $ T_\text{max} $ is the temperature resulting from the combustion. If I understand the answer correctly, this should be about 1100 K above ambient temperature, so I am currently using this term to describe the impact of thermal efficiency on fuel economy:

$$ \epsilon_\text{T} \propto \frac{1}{\eta} = \frac{T_\text{max}}{T_\text{max} - T_\text{amb}} = \frac{T_\text{amb} + 1100 \, \mathrm{K}}{1100 \, \mathrm{K}} $$

I am not sure if the increase in temperature of 1100 K is constant with altitude, so please correct me if this is wrong.


From another answer by Peter Kämpf, we know that induced drag is proportional to dynamic pressure

$$ q = \frac{v^2}{2} \cdot \rho $$

with $ v $ being the TAS and $ \rho $ the density (known from ISA). Since work required to overcome the drag per distance is proportional to the force, the fuel economy should just scale with

$$ \epsilon_\text{drag} \propto \text{TAS}^2 \cdot \rho $$

Propulsive Efficiency

From this answer we know that the propulsive efficiency for a jet engine is given by

$$ \eta_p = \frac{2}{1 + v_e / v_0} $$

where $v_e$ is the exhaust velocity and $v_0$ is the TAS. As far as I could find, there is no straight-forward way to relate $v_e$ to altitude and temperature. For the moment, I added the propulsive efficiency for a high-bypass jet engine from the following plot from Wikipedia:

Propulsive Efficiency
(image source: Wikipedia)


For the combined (relative) fuel economy term, I just multiply all previous terms:

$$ \epsilon = \epsilon_T \cdot \epsilon_\text{Drag} \cdot \epsilon_\text{Prop} $$

The following plot now shows the relative fuel required per distance. Each curve has been normalized to 1 at sea level.

Fuel Economy vs. altitude

The propulsive efficiency dominates as long as the TAS is increased. Afterwards, the lower drag dominates. The overall fuel use is almost half as low at high cruise altitudes compared to sea level.

  • $\begingroup$ Watch out. The “drag is proportional to dynamic pressure” is sloppy wording from Peter. It is a function of dynamic pressure, but there is a dynamic pressure where drag is minimal. And there is another indicated airspeed a little above that where the power of drag (drag times speed is the rate of energy loss due to drag) and these only depend on the dynamic pressure, which means indicated airspeed, and not altitude (much; see that Peter's answer). But the conclusion that drag is basically constant is correct. $\endgroup$
    – Jan Hudec
    Commented Jun 6, 2020 at 12:04
  • $\begingroup$ Your conclusion is actually correct for propeller-driven aircraft. With decent constant speed propeller the propulsive efficiency, that is fuel consumed for unit of energy actually given to the aircraft is fairly flat. And because constant drag means constant energy for travelling given distance, the fuel consumption varies little with altitude. That is, however, not the case with turbojet and turbofan engines. Their propulsive efficiency increases quite significantly with forward speed. Unfortunately I didn't find good quantification anywhere. $\endgroup$
    – Jan Hudec
    Commented Jun 6, 2020 at 12:20
  • $\begingroup$ @JanHudec Thanks for your comments. I knew I was missing at least one term. I will have a closer look at the jet propulsive efficiency... $\endgroup$
    – Bianfable
    Commented Jun 6, 2020 at 14:11
  • $\begingroup$ the problem is that good data is difficult to find. Long ago I found some NASA lectures with a Java applet that is supposed to calculate this. The four links I wrote down are: <grc.nasa.gov/WWW/k-12/airplane/specth.html>, <grc.nasa.gov/WWW/k-12/airplane/turbfan.html>, <grc.nasa.gov/WWW/k-12/airplane/ngnsim.html> and <grc.nasa.gov/WWW/K-12/airplane/EngineTheory.pdf>. Back then I didn't see sources of the applet and I didn't get around to read through the paper, but sources seem to be there now. $\endgroup$
    – Jan Hudec
    Commented Jun 6, 2020 at 18:58
  • $\begingroup$ @JanHudec I added propulsive efficiency now, but I'm not sure if the graph I found is useful for this comparison. $\endgroup$
    – Bianfable
    Commented Oct 14, 2020 at 10:02

I can not see a comprehensive answer to the first question but will answer the next three

  1. Why does fuel consumption decrease with increasing aircraft altitude?

Because of an improved ratio of true speed to total drag. See question 2

  1. What is the relation between an airplane's altitude and the drag it is experiencing?

With reduced air density more speed is required to create the same dynamic pressure. Both lift and drag are direct functions of the dynamic pressure. The limit on altitude is a combination of Angle of attack and airspeed. Absolute ceiling is determined by gross weight, maximum lift coefficient(high AoA) and maximum dynamic pressure.(most often limited by critical mach and air density) Best altitude is found near the optimum angle of attack ceiling. Which results in just under critical mach and a dynamic pressure where lift=gross weight with an AoA of best lift to drag ratio. This will be below the absolute ceiling because best L/D AoA is generally much lower than the maximum lift AoA.

  1. Why do jet engines get better fuel efficiency at high altitudes? (contains formula for thermal efficiency)I will answer

Simply they are designed for best performance at high altitude and artificially throttled at low altitude. Because they are designed to compress cold low density air at high altitude they would overheat or have mechanical stress failures if operated at maximum near sea level. The biggest limit in gas-turbine design is the thermal limits of the material used in the high turbine first stage.(strength at a temperature) Operating below maximum is less efficient, this is a common thermodynamics/entropy engineering problem, the larger the difference in energy the more efficiently that energy can be harnessed(converted to another form) or transferred in or out of a system. This is also a limit in steam engine design, higher temperature boilers and higher pressure steam make them more efficient but material properties and safety are the limits with common heat sources.

  • $\begingroup$ Climb and decent roughly cancel out. The climb adds potential energy and the decent converts potential energy into kinetic energy. Any difference comes down to individual aircraft/engine design and operation parameters, and so is difficult to generalize. $\endgroup$
    – Max Power
    Commented Oct 14, 2020 at 2:44
  • $\begingroup$ The ratio of true speed to total drag is, in itself, irrelevant. Since the drag ends up being the same, the work for the same distance also remains the same (because work is force times distance). If the efficiency of the engine is independent of speed, the fuel consumption is mostly independent of altitude, which is actually true of piston-powered propeller planes! Turbojet and turbofan engines, and to some extent even turbofan engines are however more efficient at higher speed (the ram pressure increases their effective compression ratio), and that is the reason. $\endgroup$
    – Jan Hudec
    Commented Oct 14, 2020 at 7:30
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    $\begingroup$ @MaxPower Thanks for the answer, but I think you misunderstood my actual question. The "questions" you answered are other related questions, which already have answers. I was interested in how to combine things quantitatively. $\endgroup$
    – Bianfable
    Commented Oct 14, 2020 at 9:50

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